# Prove that the Frobenius norm is invariant under orthonormal projection

Assume I can express a rank-deficient, $$N\times N$$ symmetric covariance matrix $$\Sigma$$ as

$$$$\Sigma=\mathbf{USU}^\top$$$$

where $$\mathbf{U}$$ is an $$L\times N$$ orthonormal matrix, i.e. $$\mathbf{U^\top U}=\mathbf{I}_L$$.

I am wondering whether the Frobenius norm $$|\Sigma|$$ is equal to $$|\mathbf{S}|$$. I can see how this is true for the special case where $$\mathbf{U}$$ is an orthonormalizing (eigenvector) basis (so that $$\mathbf{S}$$ is diagonal), and haven't been able to show otherwise in numerical experiments, but I'm not sure how to evaluate more generally.

The answer is yes. Note that \begin{align} |\Sigma|^2 &= |USU^T|^2 = \operatorname{tr}[(USU^T)(USU^T)^T] \\ & = \operatorname{tr}[US(U^TU)S^TU^T] \\ & = \operatorname{tr}[U(SS^TU^T)] = \operatorname{tr}[(SS^TU^T)U] \\ & = \operatorname{tr}[SS^T(U^TU)] = \operatorname{tr}[SS^T] = |S|^2. \end{align}
Alternatively, complete $$U$$ to a square orthogonal matrix $$V=\pmatrix{U&\ast}$$. Then $$\Sigma=USU^T=V\pmatrix{S\\ &0}V^T.$$ Therefore $$\|\Sigma\|_F =\left\|V\pmatrix{S\\ &0}V^T\right\|_F =\left\|\pmatrix{S\\ &0}\right\|_F=\|S\|_F$$.